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MAXImum [283]
3 years ago
14

Solve problem using substitution a = 3b + 1 5b - 2a = 1

Mathematics
2 answers:
Lorico [155]3 years ago
6 0
Hope this helps you !!!
vampirchik [111]3 years ago
5 0
Solve for the first variable in one of the equations, then substitute the result into the other equation.

a=−13, b=−5
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Is (3,8) a solution to this system of eqautions?<br> y=7x+8<br> y=x+1<br> Yes or No?
erastova [34]

Answer:

Step-by-step explanation:

plug the numbers in and see if it is true

x=3, y=8

y=7x+8

8=7(3)+8

8=29

is not true, so it is not a solution. even if it might be a solution for the other equation, it is not true for this equation so it is not a solution for this system

how to find an equation: since y=7x+8 and y=x+1, then

7x+8=y, and y=x+1

7x+8=y=y=x+1

7x+8=x+1

6x+8=1

6x=-7

x=-7/6

y=x+1

y=-1/6

solution: (-1 1/6, -1/6)

6 0
2 years ago
C + G + I +F is the formula to determine GDP. What does the G in the formula represent?
jenyasd209 [6]
"G" would have to stand for government spending, so the answer is (a).
6 0
2 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
2 years ago
Multiply. Write your answer in standard form<br> 3i(-5+i)
ipn [44]
-14i is the answer to the equation
8 0
3 years ago
Can y’all help me please? :))
german

Answer:

3.75

Step-by-step explanation:

1.25x3=3.75 easy. you just need to times the height by length

6 0
3 years ago
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